3 - Chemical Equations and Reaction Stoichiometry

3

Chemical Equations and Reaction Stoichiometry

OUTLINE 3-1 Chemical Equations 3-2 Calculations Based on Chemical Equations 3-3 The Limiting Reactant Concept 3-4 Percent Yields from Chemical Reactions

3-5 3-6 3-7 3-8

Sequential Reactions Concentrations of Solutions Dilution of Solutions Using Solutions in Chemical Reactions

OBJECTIVES After you have studied this chapter, you should be able to •

Write balanced chemical equations to describe chemical reactions

•

Interpret balanced chemical equations to calculate the moles of reactants and products involved in each of the reactions

•

Interpret balanced chemical equations to calculate the masses of reactants and products involved in each of the reactions

•

Determine which reactant is the limiting reactant in reactions

•

Use the limiting reactant concept in calculations recording chemical equations

•

Compare the amount of substance actually formed in a reaction (actual yield) with the predicted amount (theoretical yield), and determine the percent yield

•

Work with sequential reactions

•

Use the terminology of solutions — solute, solvent, concentration

•

Calculate concentrations of solutions when they are diluted

•

Carry out calculations related to the use of solutions in chemical reactions

I

n Chapter 2 we studied composition stoichiometry, the quantitative relationships among elements in compounds. In this chapter as we study reaction stoichiometry — the quantitative relationships among substances as they participate in chemical reactions — we ask several important questions. How can we describe the reaction of one substance with another? How much of one substance reacts with a given amount of another substance? Which reactant determines the amounts of products formed in a chemical reaction? How can we describe reactions in aqueous solutions? Whether we are concerned with describing a reaction used in a chemical analysis, one used industrially in the production of a plastic, or one that occurs during metabolism in the body, we must describe it accurately. Chemical equations represent a very precise, yet a very versatile, language that describes chemical changes. We begin our study by examining chemical equations.

Methane, CH4, is the main component of natural gas.

Steel wool burns in pure oxygen gas. 4Fe ⫹ 3O2 88n 2Fe2O3

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CHAPTER 3: Chemical Equations and Reaction Stoichiometry

3-1 CHEMICAL EQUATIONS

Sometimes it is not possible to represent a chemical change with a single chemical equation. For example, when too little O2 is present, both CO2 and CO are found as products, and a second chemical equation must be used to describe the process. In the present case (excess oxygen), only one equation is required. The arrow may be read “yields.” The capital Greek letter delta (⌬) is sometimes used in place of the word “heat.”

Chemical reactions always involve changing one or more substances into one or more different substances. In other words, chemical reactions rearrange atoms or ions to form other substances. Chemical equations are used to describe chemical reactions, and they show (1) the substances that react, called reactants; (2) the substances formed, called products; and (3) the relative amounts of the substances involved. We write the reactants to the left of an arrow and the products to the right of the arrow. As a typical example, let’s consider the combustion (burning) of natural gas, a reaction used to heat buildings and cook food. Natural gas is a mixture of several substances, but the principal component is methane, CH4. The equation that describes the reaction of methane with excess oxygen is CH4 ⫹ 2O2 88n CO2 ⫹ 2H2O

      

General Chemistry CD-ROM, Screens 4.1–4.4, Equations.

      

See the Saunders Interactive

reactants

products

What does this equation tell us? In the simplest terms, it tells us that methane reacts with oxygen to produce carbon dioxide, CO2, and water. More specifically, it says that for every CH4 molecule that reacts, two molecules of O2 also react, and that one CO2 molecule and two H2O molecules are formed. That is, CH4 1 molecule

⫹

2O2 2 molecules

heat

8888n

⫹

CO2 1 molecule

2H2O 2 molecules

This description of the reaction of CH4 with O2 is based on experimental observations. By this we mean experiments have shown that when one CH4 molecule reacts with two O2 molecules, one CO2 molecule and two H2O molecules are formed. Chemical equations are based on experimental observations. Special conditions required for some reactions are indicated by notation over the arrow. Figure 3-1 is a pictorial representation of the rearrangement of atoms described by this equation. As we pointed out in Section 1-1, there is no detectable change in the quantity of matter during an ordinary chemical reaction. This guiding principle, the Law of Conservation of Matter, provides the basis for “balancing” chemical equations and for calculations based on those equations. Because matter is neither created nor destroyed during a chemical reaction, a balanced chemical equation must always include the same number of each kind of atom on both sides of the equation.

O

Figure 3-1 Two representations of the reaction of methane with oxygen to form carbon dioxide and water. Chemical bonds are broken and new ones are formed in each representation. Part (a) illustrates the reaction using models, and (b) uses chemical formulas.

H C

(a) H

O

H

O

⫹

O

C

O

⫹

H O

O

O H

(b)

H

H

CH4

⫹

2O2

CO2

⫹

H

2H2O

3-1 Chemical Equations

Chemists usually write equations with the smallest possible whole-number coefficients. Before we attempt to balance an equation, all substances must be represented by formulas that describe them as they exist. For instance, we must write H2 to represent diatomic hydrogen molecules — not H, which represents hydrogen atoms. Once the correct formulas are written, the subscripts in the formulas may not be changed. Different subscripts in formulas specify different compounds, so changing the formulas would mean that the equation would no longer describe the same reaction. Dimethyl ether, C2H6O, burns in an excess of oxygen to give carbon dioxide and water. Let’s balance the equation for this reaction. In unbalanced form, the equation is C2H6O ⫹ O2 88n CO2 ⫹ H2O Carbon appears in only one compound on each side, and the same is true for hydrogen. We begin by balancing these elements: C2H6O ⫹ O2 88n 2CO2 ⫹ 3H2O Now we have an odd number of atoms of O on each side. The single O in C2H6O balances one of the atoms of O on the right. We balance the other six by placing a coefficient of 3 before O2 on the left. C2H6O ⫹ 3O2 88n 2CO2 ⫹ 3H2O When we think we have finished the balancing, we should always do a complete check for each element, as shown in red in the margin. Let’s generate the balanced equation for the reaction of aluminum metal with hydrochloric acid to produce aluminum chloride and hydrogen. The unbalanced “equation” is Al ⫹ HCl 88n AlCl3 ⫹ H2 As it now stands, the “equation” does not satisfy the Law of Conservation of Matter because there are two H atoms in the H2 molecule and three Cl atoms in one formula unit of AlCl3 (right side), but only one H atom and one Cl atom in the HCl molecule (left side). Let us first balance chlorine by putting a coefficient of 3 in front of HCl. Al ⫹ 3HCl 88n AlCl3 ⫹ H2 Now there are 3H on the left and 2H on the right. The least common multiple of 3 and 2 is 6; to balance H, we multiply the 3HCl by 2 and the H2 by 3. Al ⫹ 6HCl 88n AlCl3 ⫹ 3H2 Now Cl is again unbalanced (6Cl on the left, 3 on the right), but we can fix this by putting a coefficient of 2 in front of AlCl3 on the right. Al ⫹ 6HCl 88n 2AlCl3 ⫹ 3H2 Now all elements except Al are balanced (1 on the left, 2 on the right); we complete the balancing by putting a coefficient of 2 in front of Al on the left. 2Al aluminum

⫹

6HCl hydrochloric acid

88n

2AlCl3 aluminum chloride

⫹

3H2 hydrogen

Balancing chemical equations “by inspection” is a trial-and-error approach. It requires a great deal of practice, but it is very important! Remember that we use the smallest wholenumber coefficients. Some chemical equations are difficult to balance by inspection or “trial and error.” In Chapter 11 we will learn methods for balancing complex equations.

91

In Reactants

In Products

2C, 6H, 3O 1C, 2H, 3O C, H are not balanced In Reactants

In Products

2C, 6H, 3O 2C, 6H, 7O Now O is not balanced

In Reactants

In Products

2C, 6H, 7O 2C, 6H, 7O Now the equation is balanced

In Reactants

In Products

1Al, 1H, 1Cl 1Al, 2H, 3Cl H, Cl are not balanced

In Reactants

In Products

1Al, 3H, 3Cl 1Al, 2H, 3Cl Now H is not balanced In Reactants

In Products

1Al, 6H, 6Cl 1Al, 6H, 3Cl Now Cl is not balanced In Reactants

In Products

1Al, 6H, 6Cl 2Al, 6H, 6Cl Now Al is not balanced In Reactants

In Products

2Al, 6H, 6Cl 2Al, 6H, 6Cl Now the equation is balanced

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Problem-Solving Tip: Balancing Chemical Equations There is no one best place to start when balancing a chemical equation, but the following suggestions might be helpful: (1) Look for elements that appear in only one place on each side of the equation (in only one reactant and in only one product), and balance those elements first. (2) If free, uncombined elements appear on either side, balance them last. Notice how these suggestions worked in the procedures illustrated in this section. Above all, remember that we should never change subscripts in formulas, because doing so would describe different substances. We only adjust the coefficients to balance the equation.

3-2 CALCULATIONS BASED ON CHEMICAL EQUATIONS

See the Saunders Interactive General Chemistry CD-ROM, Screens 5.1–5.3, Stoichiometry.

We are now ready to use chemical equations to calculate the relative amounts of substances involved in chemical reactions. Let us again consider the combustion of methane in excess oxygen. The balanced chemical equation for that reaction is CH4 ⫹ 2O2 88n CO2 ⫹ 2H2O On a quantitative basis, at the molecular level, the equation says CH4

⫹

1 molecule of methane

2O2

CO2

88n

2 molecules of oxygen

1 molecule of carbon dioxide

⫹ 2H2O 2 molecules of water

EXAMPLE 3-1 Number of Molecules How many O2 molecules react with 47 CH4 molecules according to the preceding equation? A balanced chemical equation may be interpreted on a molecular basis.

Plan The balanced equation tells us that one CH4 molecule reacts with two O2 molecules. We can construct two unit factors from this fact: 1 CH4 molecule ᎏᎏ 2 O2 molecules

and

2 O2 molecules ᎏᎏ 1 CH4 molecule

These expressions are unit factors for this reaction because the numerator and denominator are chemically equivalent. In other words, the numerator and the denominator represent the same amount of reaction. To convert CH4 molecules to O2 molecules, we multiply by the second of the two factors. Solution 2 O2 molecules __ ? O2 molecules ⫽ 47 CH4 molecules ⫻ ᎏᎏ ⫽ 94 O2 molecules 1 CH4 molecule You should now work Exercise 8. We usually cannot work with individual molecules; a mole of a substance is an amount we might use in a laboratory experiment.

A chemical equation also indicates the relative amounts of each reactant and product in a given chemical reaction. We showed earlier that formulas can represent moles of sub-

3-2 Calculations Based On Chemical Equations

93

stances. Suppose Avogadro’s number of CH4 molecules, rather than just one CH4 molecule, undergo this reaction. Then the equation can be written CH4

⫹

6.02 ⫻ 1023 molecules ⫽ 1 mol

2O2

CO2

88n

2(6.02 ⫻ 1023 molecules) ⫽ 2 mol

6.02 ⫻ 1023 molecules ⫽ 1 mol

⫹

2H2O

2(6.02 ⫻ 1023molecules) ⫽ 2 mol

This interpretation tells us that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.

A balanced chemical equation may be interpreted in terms of moles of reactants and products.

Problem-Solving Tip: Are you Using a Balanced Equation? Never start a calculation involving a chemical reaction without first checking that the equation is balanced.

EXAMPLE 3-2 Number of Moles Formed How many moles of water could be produced by the reaction of 3.5 mol of methane with excess oxygen (i.e., more than a sufficient amount of oxygen is present)? Plan The equation for the combustion of methane CH4 ⫹ 2O2 88n CO2 ⫹ 2H2O 1 mol

2 mol

1 mol

2 mol

shows that 1 mol of methane reacts with 2 mol of oxygen to produce 2 mol of water. From this information we construct two unit factors: 1 mol CH4 ᎏᎏ 2 mol H2O

and

2 mol H2O ᎏᎏ 1 mol CH4

Please don’t try to memorize unit factors for chemical reactions; rather, learn the general method for constructing them from balanced chemical equations.

We use the second factor in this calculation. Solution 2 mol H2O __ ? mol H2O ⫽ 3.5 mol CH4 ⫻ ᎏᎏ ⫽ 7.0 mol H2O 1 mol CH4 You should now work Exercises 10 and 14.

We know the mass of 1 mol of each of these substances, so we can also write CH4 ⫹

2O2

88n CO2 ⫹ 2H2O 1 mol 44.0 g 44.0 g

2 mol 2(18.0 g) 36.0 g

        

2 mol 2(32.0 g) 64.0 g

        

1 mol 16.0 g 16.0 g

80.0 g reactants

80.0 g products

The equation now tells us that 16.0 grams of CH4 reacts with 64.0 grams of O2 to form 44.0 grams of CO2 and 36.0 grams of H2O. The Law of Conservation of Matter is satisfied. Chemical equations describe reaction ratios, that is, the mole ratios of reactants and products as well as the relative masses of reactants and products.

A balanced equation may be interpreted on a mass basis.

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CHAPTER 3: Chemical Equations and Reaction Stoichiometry

Problem-Solving Tip: Use the Mole Ratio in Calculations with Balanced Chemical Equations The most important way of interpretating the balanced chemical equation is in terms of moles. We use the coefficients to get the mole ratio of any two substances we want to relate. Then we apply it as moles of mole ratio   moles of   desired ⫽  substance  ⫻  from balanced   chemical equation   given  substance It is important to include the substance formulas as part of the units; this can help us decide how to set up the unit factors. Notice that in Example 3-2, we want to cancel the term mol CH4, so we know that mol CH4 must be in the denominator of the mole ratio by which we multiply; we want mol O2 as the answer, so mol O2 must appear mol in the numerator of the mole ratio. In other words, do not just write ᎏᎏ; write mol mol of something ᎏᎏᎏ, giving the formulas of the two substances involved. mol of something else

EXAMPLE 3-3 Mass of a Reactant Required What mass of oxygen is required to react completely with 1.20 mol of CH4? Plan The balanced equation CH4 ⫹ 1 mol 16.0 g

2O2

88n CO2 ⫹ 2H2O

2 mol 2(32.0 g)

1 mol 44.0 g

2 mol 2(18.0 g)

gives the relationships among moles and grams of reactants and products. mol CH4 88n mol O2 88n g O2 Solution 2 mol O2 32.0 g O2 __ ? g O2 ⫽ 1.20 mol CH4 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 76.8 g O2 1 mol CH4 1 mol O2

EXAMPLE 3-4 Mass of a Reactant Required What mass of oxygen is required to react completely with 24.0 g of CH4? Plan Recall the balanced equation in Example 3-3. CH4 ⫹ 1 mol 16.0 g

2O2 2 mol 2(32.0) g

88n CO2 ⫹ 2H2O 1 mol 44.0 g

2 mol 2(18.0) g

3-2 Calculations Based On Chemical Equations

95

This shows that 1 mol of CH4 reacts with 2 mol of O2. These two quantities are chemically equivalent, so we can construct unit factors. Solution CH4 ⫹ 2O2 88n CO2 ⫹ 2H2O 1 mol

2 mol

1 mol

2 mol

1 mol CH4 __ ? mol CH4 ⫽ 24.0 g CH4 ⫻ ᎏᎏ ⫽ 1.50 mol CH4 16.0 g CH4 XXXX 6 XXX

6 gXXXXXXXXXXXXXXX 2 mol O 2

__ ? mol O2 ⫽ 1.50 mol CH4 ⫻ ᎏᎏ ⫽ 3.00 mol O2 1 mol CH4 XXXXXX 6

6 gXXXXXXXXXXXXXXXX 32.0 g O 2

__ ? g O2 ⫽ 3.00 mol O2 ⫻ ᎏᎏ ⫽ 96.0 g O2 1 mol O2

Here we solve the problem in three steps and so we convert 1. g CH4 n mol CH4 2. mol CH4 n mol O2 3. mol O2 n g O2

All these steps could be combined into one setup in which we convert g of CH4 88n mol of CH4 88n mol of O2 88n g of O2 1 mol CH4 2 mol O2 32.0 g O2 __ ? g O2 ⫽ 24.0 g CH4 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 96.0 g O2 16.0 g CH4 1 mol CH4 1 mol O2 The same answer, 96.0 g of O2, is obtained by both methods. You should now work Exercise 18.

The question posed in Example 3-4 may be reversed, as in Example 3-5.

EXAMPLE 3-5 Mass of a Reactant Required What mass of CH4, in grams, is required to react with 96.0 grams of O2? Plan We recall that one mole of CH4 reacts with two moles of O2. Solution 16.0 g CH4 1 mol O2 1 mol CH4 __ ? g CH4 ⫽ 96.0 g O2 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 24.0 g CH4 32.0 g O2 2 mol O2 1 mol CH4

These unit factors are the reciprocals of those used in Example 3-4.

You should now work Exercise 22.

This is the amount of CH4 in Example 3-4 that reacted with 96.0 grams of O2.

EXAMPLE 3-6 Mass of a Product Formed Most combustion reactions occur in excess O2, that is, more than enough O2 to burn the substance completely. Calculate the mass of CO2, in grams, that can be produced by burning 6.00 mol of CH4 in excess O2.

It is important to recognize that the reaction must stop when the 6.0 mol of CH4 has been used up. Some O2 will remain unreacted.

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CHAPTER 3: Chemical Equations and Reaction Stoichiometry

Plan The balanced equation tells us that one mole of CH4 produces one mole of CO2. CH4 ⫹ 1 mol 16.0 g

88n CO2 ⫹ 2H2O

2O2 2 mol 2(32.0 g)

1 mol 44.0 g

2 mol 2(18.0 g)

Solution 1 mol CO2 44.0 g CO2 __ ? g CO2 ⫽ 6.00 mol CH4 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 2.64 ⫻ 102 g CO2 1 mol CH4 1 mol CO2 You should now work Exercise 20.

Reaction stoichiometry usually involves interpreting a balanced chemical equation to relate a given bit of information to the desired bit of information.

EXAMPLE 3-7 Mass of a Reactant Required Phosphorus, P4, burns with excess oxygen to form tetraphosphorus decoxide, P4O10. In this reaction, what mass of P4 reacts with 1.50 moles of O2? Plan The balanced equation tells us that one mole of P4 reacts with five moles of O2. P4 ⫹ 5O2 88n P4O10 1 mol

5 mol

1 mol

mol O2 88n mol P4 88n mass P4 Solution 1 mol P4 124.0 g P4 __ ? g P4 ⫽ 1.50 mol O2 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 37.2 g P4 5 mol O2 mol P4 You should now work Exercise 24.

The possibilities for this kind of problem solving are limitless. Before you continue, you should work Exercises 8–25 at the end of the chapter.

3-3 THE LIMITING REACTANT CONCEPT

See the Saunders Interactive General Chemistry CD-ROM, Screen 5.5, Limiting Reactants.

In the problems we have worked thus far, the presence of an excess of one reactant was stated or implied. The calculations were based on the substance that was used up first, called the limiting reactant. Before we study the concept of the limiting reactant in stoichiometry, let’s develop the basic idea by considering a simple but analogous nonchemical example. Suppose you have four slices of ham and six slices of bread and you wish to make as many ham sandwiches as possible using only one slice of ham and two slices of bread per sandwich. Obviously, you can make only three sandwiches, at which point you run out of

3-3 The Limiting Reactant Concept

bread. (In a chemical reaction this would correspond to one of the reactants being used up — so the reaction would stop.) The bread is therefore the “limiting reactant,” and the extra slice of ham is the “excess reactant.” The amount of product, ham sandwiches, is determined by the amount of the limiting reactant, bread in this case. The limiting reactant is not necessarily the reactant present in the smallest amount. We have four slices of ham, the smallest amount, and six slices of bread, but the “reaction ratio” is two slices of bread to one piece of ham, and so bread is the limiting reactant.

EXAMPLE 3-8 Limiting Reactant What mass of CO2 could be formed by the reaction of 16.0 g of CH4 with 48.0 g of O2? Plan The balanced equation tells us that one mole of CH4 reacts with two moles of O2. CH4 ⫹ 1 mol 16.0 g

2O2 2 mol 2(32.0 g)

88n CO2 ⫹ 2H2O 1 mol 44.0 g

2 mol 2(18.0 g)

We are given masses of both CH4 and O2, so we calculate the number of moles of each reactant, and then determine the number of moles of each reactant required to react with the other. From these calculations we identify the limiting reactant. Then we base the calculation on it. Solution 1 mol CH4 __ ? mol CH4 ⫽ 16.0 g CH4 ⫻ ᎏᎏ ⫽ 1.00 mol CH4 16.0 g CH4 1 mol O2 __ ? mol O2 ⫽ 48.0 g O2 ⫻ ᎏᎏ ⫽ 1.50 mol O2 32.0 g O2 Now we return to the balanced equation. First we calculate the number of moles of O2 that would be required to react with 1.00 mole of CH4. 2 mol O2 __ ? mol O2 ⫽ 1.00 mol CH4 ⫻ ᎏᎏ ⫽ 2.00 mol O2 1 mol CH4 We see that 2.00 moles of O2 is required, but we have 1.50 moles of O2, so O2 is the limiting reactant. Alternatively, we can calculate the number of moles of CH4 that would react with 1.50 moles of O2. 1 mol CH4 __ ? mol CH4 ⫽ 1.50 mol O2 ⫻ ᎏᎏ ⫽ 0.750 mol CH4 2 mol O2 This tells us that only 0.750 mole of CH4 would be required to react with 1.50 moles of O2. But we have 1.00 mole of CH4, so we see again that O2 is the limiting reactant. The reaction must stop when the limiting reactant, O2, is used up, so we base the calculation on O2. g of O2 88n mol of O2 88n mol of CO2 88n g of CO2 44.0 g CO2 1 mol O2 1 mol CO2 __ ? g CO2 ⫽ 48.0 g O2 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 33.0 g CO2 32.0 O2 2 mol O2 1 mol CO2 You should now work Exercise 26.

97

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CHAPTER 3: Chemical Equations and Reaction Stoichiometry

Thus, 33.0 grams of CO2 is the most CO2 that can be produced from 16.0 grams of CH4 and 48.0 grams of O2. If we had based our calculation on CH4 rather than O2, our answer would be too big (44.0 grams) and wrong because more O2 than we have would be required. Another approach to problems like Example 3-8 is to calculate the number of moles of each reactant: 1 mol CH4 __ ? mol CH4 ⫽ 16.0 g CH4 ⫻ ᎏᎏ ⫽ 1.00 mol CH4 16.0 g CH4 1 mol O2 __ ? mol O2 ⫽ 48.0 g O2 ⫻ ᎏᎏ ⫽ 1.50 mol O2 32.0 g O2 Then we return to the balanced equation. We first calculate the required ratio of reactants as indicated by the balanced chemical equation. We then calculate the available ratio of reactants and compare the two: Required Ratio 1 mol CH4 0.500 mol CH4 ᎏᎏ ⫽ ᎏᎏ 2 mol O2 1.00 mol O2

Available Ratio 1.00 mol CH4 0.667 mol CH4 ᎏᎏ ⫽ ᎏᎏ 1.50 mol O2 1.00 mol O2

We see that each mole of O2 requires exactly 0.500 mol of CH4 to be completely used up. We have 0.667 mol of CH4 for each mole of O2, so there is more than enough CH4 to react with the O2 present. That means that there is insufficient O2 to react with all of the available CH4. The reaction must stop when the O2 is gone; O2 is the limiting reactant, and we must base the calculation on it. (If the available ratio of CH4 to O2 had been smaller than the required ratio, we would have concluded that there is not enough CH4 to react with all of the O2, and CH4 would have been the limiting reactant.)

Problem-Solving Tip: Choosing the Limiting Reactant Students often wonder how to know which ratio to calculate to help find the limiting reactant. (1) The ratio must involve the two reactants with amounts given in the problem. (2) It doesn’t matter which way you calculate the ratio, as long as you calculate both required ratio and available ratio in the same order. For example, we could mol O2 calculate the required and available ratio of ᎏᎏ in the approach we just mol CH4 illustrated. If you can’t decide how to solve a limiting reactant problem, as a last resort, do the entire calculation twice — once based on each reactant amount given. The smaller answer is the right one.

EXAMPLE 3-9 Limiting Reactant What is the maximum mass of Ni(OH)2 that could be prepared by mixing two solutions that contain 25.9 g of NiCl2 and 10.0 g of NaOH, respectively? NiCl2 ⫹ 2NaOH 88n Ni(OH)2 ⫹ 2NaCl

3-4 Percent Yields from Chemical Reactions

99

Plan Interpreting the balanced equation as usual, we have NiCl2 ⫹ 2NaOH 88n Ni(OH)2 ⫹ 2NaCl 1 mol 129.6 g

2 mol 2(40.0 g)

1 mol 92.7 g

2 mol 2(58.4 g)

We determine the number of moles of NiCl2 and NaOH present. Then we find the number of moles of each reactant required to react with the other reactant. These calculations identify the limiting reactant. We base the calculation on it. Solution 1 mol NiCl2 __ ? mol NiCl2 ⫽ 25.9 g NiCl2 ⫻ ᎏᎏ ⫽ 0.200 mol NiCl2 129.6 g NiCl2 1 mol NaOH __ ? mol NaOH ⫽ 10.0 g NaOH ⫻ ᎏᎏ ⫽ 0.250 mol NaOH 40.0 g NaOH

Even though the reaction occurs in aqueous solution, this calculation is similar to earlier examples because we are given the amounts of pure reactants.

We return to the balanced equation and calculate the number of moles of NaOH required to react with 0.200 mol of NiCl2. 2 mol NaOH __ ? mol NaOH ⫽ 0.200 mol NiCl2 ⫻ ᎏᎏ ⫽ 0.400 mol NaOH 1 mol NiCl2 But we have only 0.250 mol of NaOH, so NaOH is the limiting reactant. g of NaOH 88n mol of NaOH 88n mol Ni(OH)2 88n g of Ni(OH)2 1 mol NaOH 1 mol Ni(OH)2 92.7 g Ni(OH)2 __ ? g Ni(OH)2 ⫽ 10.0 g NaOH ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ 40.0 g NaOH 2 mol NaOH 1 mol Ni(OH)2 ⫽ 11.6 g Ni(OH)2 You should now work Exercises 30 and 34.

Problem-Solving Tip: How Can We Recognize a Limiting Reactant Problem? When we are given the amounts of two (or more) reactants, we should suspect that we are dealing with a limiting reactant problem. It is very unlikely that exactly the stoichiometric amounts of both reactants are present in a reaction mixture.

A precipitate of solid Ni(OH)2 forms when colorless NaOH solution is added to green NiCl2 solution.

3-4 PERCENT YIELDS FROM CHEMICAL REACTIONS The theoretical yield from a chemical reaction is the yield calculated by assuming that the reaction goes to completion. In practice we often do not obtain as much product from a reaction mixture as is theoretically possible. This is true for several reasons. (1) Many reactions do not go to completion; that is, the reactants are not completely converted to products. (2) In some cases, a particular set of reactants undergoes two or more reactions

In the examples we have worked to this point, the amounts of products that we calculated were theoretical yields.

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CHAPTER 3: Chemical Equations and Reaction Stoichiometry

simultaneously, forming undesired products as well as desired products. Reactions other than the desired one are called “side reactions.” (3) In some cases, separation of the desired product from the reaction mixture is so difficult that not all of the product formed is successfully isolated. The actual yield is the amount of a specified pure product actually obtained from a given reaction. The term percent yield is used to indicate how much of a desired product is obtained from a reaction.

See the Saunders Interactive General Chemistry CD-ROM, Screen 5.6, Percent Yield.

actual yield of product percent yield ⫽ ᎏᎏᎏ ⫻ 100% theoretical yield of product Consider the preparation of nitrobenzene, C6H5NO2, by the reaction of a limited amount of benzene, C6H6, with excess nitric acid, HNO3. The balanced equation for the reaction may be written as C6H6 ⫹ HNO3 88n C6H5NO2 ⫹ H2O 1 mol 78.1 g

1 mol 63.0 g

1 mol 123.1 g

1 mol 18.0 g

EXAMPLE 3-10 Percent Yield A 15.6-gram sample of C6H6 is mixed with excess HNO3. We isolate 18.0 grams of C6H5NO2. What is the percent yield of C6H5NO2 in this reaction? Plan First we interpret the balanced chemical equation to calculate the theoretical yield of C6H5NO2. Then we use the actual (isolated) yield and the previous definition to calculate the percent yield. Solution We calculate the theoretical yield of C6H5NO2. It is not necessary to know the mass of one mole of HNO3 to solve this problem.

1 mol C6H6 1 mol C6H5NO2 123.1 g C6H5NO2 __ ? g C6H5NO2 ⫽ 15.6 g C6H6 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏᎏ 78.1 g C6H6 1 mol C6H6 1 mol C6H5NO2 ⫽ 24.6 g C6H5NO2 m theoretical yield This tells us that if all the C6H6 were converted to C6H5NO2 and isolated, we should obtain 24.6 grams of C6H5NO2 (100% yield). We isolate only 18.0 grams of C6H5NO2, however. actual yield of product 18.0 g percent yield ⫽ ᎏᎏᎏ ⫻ 100% ⫽ ᎏ ⫻ 100% theoretical yield of product 24.6 g ⫽ 73.2 percent yield You should now work Exercise 38.

The amount of nitrobenzene obtained in this experiment is 73.2% of the amount that would be expected if the reaction had gone to completion, if there were no side reactions, and if we could have recovered all of the product as pure substance.

3-5 Sequential Reactions

3-5 SEQUENTIAL REACTIONS Often more than one reaction is required to change starting materials into the desired product. This is true for many reactions that we carry out in the laboratory and for many industrial processes. These are called sequential reactions. The amount of desired product from each reaction is taken as the starting material for the next reaction.

EXAMPLE 3-11 Sequential Reactions At high temperatures, carbon reacts with water to produce a mixture of carbon monoxide, CO, and hydrogen, H2. heat

C ⫹ H2O 8888n CO ⫹ H2 Carbon monoxide is separated from H2 and then used to separate nickel from cobalt by forming a gaseous compound, nickel tetracarbonyl, Ni(CO)4. Ni ⫹ 4CO 88n Ni(CO)4 What mass of Ni(CO)4 could be obtained from the CO produced by the reaction of 75.0 g of carbon? Assume 100% yield. Plan We interpret both chemical equations in the usual way, and solve the problem in two steps. They tell us that one mole of C produces one mole of CO and that four moles of CO is required to produce one mole of Ni(CO)4. 1. We determine the number of moles of CO formed in the first reaction. 2. From the number of moles of CO produced in the first reaction, we calculate the number of grams of Ni(CO)4 formed in the second reaction. Solution 1.

C

⫹ H2O 88n CO ⫹ H2

1 mol 12.0 g

1 mol

1 mol

1 mol

1 mol C 1 mol CO __ ? mol CO ⫽ 75.0 g C ⫻ ᎏ ⫻ ᎏᎏ ⫽ 6.25 mol CO 12.0 g C 1 mol C Ni ⫹ 4CO 88n Ni(CO)4

2.

1 mol

4 mol

1 mol 171 g

1 mol Ni(CO)4 171 g Ni(CO)4 __ ? g Ni(CO)4 ⫽ 6.25 mol CO ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 267 g Ni(CO)4 4 mol CO 1 mol Ni(CO)4 Alternatively, we can set up a series of unit factors based on the conversions in the reaction sequence and solve the problem in one setup. g C 88n mol C 88n mol CO 88n mol Ni(CO)4 88n g Ni(CO)4 1 mol C 1 mol CO 1 mol Ni(CO)4 171 g Ni(CO)4 __ ? g Ni(CO)4 ⫽ 75.0 g C ⫻ ᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 267 g Ni(CO)4 12.0 g C 1 mol C 4 mol CO 1 mol Ni(CO)4 You should now work Exercise 46.

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EXAMPLE 3-12 Sequential Reactions Two large uses for H3PO4 are in fertilizers and cola drinks. Approximately 100 lb of H3PO4– based fertilizer are used per year per person in America.

Phosphoric acid, H3PO4, is a very important compound used to make fertilizers. It is also present in cola drinks. H3PO4 can be prepared in a two-step process. Rxn 1:

P4 ⫹ 5O2 88n P4O10

Rxn 2:

P4O10 ⫹ 6H2O 88n 4H3PO4

We allow 272 grams of phosphorus to react with excess oxygen, which forms tetraphosphorus decoxide, P4O10, in 89.5% yield. In the second step reaction, a 96.8% yield of H3PO4 is obtained. What mass of H3PO4 is obtained? Plan 1. We interpret the first equation as usual and calculate the amount of P4O10 obtained. P4 ⫹ 1 mol 124 g

5O2

88n P4O10

5 mol 5(32.0 g)

1 mol 284 g

g P4 88n mol P4 88n mol P4O10 88n g P4O10 2. Then we interpret the second equation and calculate the amount of H3PO4 obtained from the P4O10 from the first step. P4O10 ⫹ 6H2O 88n 4H3PO4 1 mol 284 g

6 mol 6(18.0 g)

4 mol 4(98.0 g)

g P4O10 88n mol P4O10 88n mol H3PO4 88n g H3PO4 Solution 284 g P4O10 theoretical 1 mol P4 1 mol P4O10 1. __ ? g P4O10 ⫽ 272 g P4 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏᎏ 124 g P4 1 mol P4 1 mol P4O10 theoretical The unit factors that account for less than 100% reaction and less than 100% recovery are included in parentheses.

89.5 g P4O10 actual ⫻ ᎏᎏᎏ ⫽ 558 g P4O10 100 g P4O10 theoretical

冢

冣

4 mol H3PO4 1 mol P4O10 2. __ ? g H3PO4 ⫽ 558 g P4O10 ⫻ ᎏᎏ ⫻ ᎏᎏ 284 g P4O10 1 mol P4O10 98.0 g H3PO4 theoretical 96.8 g H3PO4 actual ⫻ ᎏᎏᎏ ⫻ ᎏᎏᎏ ⫽ 746 g H3PO4 1 mol H3PO4 theoretical 100 g H3PO4 theoretical

冢

冣

You should now work Exercises 48 and 50.

Chemists have determined the structures of many naturally occurring compounds. One way of proving the structure of such a compound is by synthesizing it from available starting materials. Professor Grieco, now at Indiana University, was assisted by Majetich and Ohfune in the synthesis of helenalin, a powerful anticancer drug, in a 40-step process. This 40-step synthesis gave a remarkable average yield of about 90% for each step, which still resulted in an overall yield of only about 1.5%.

3-6 Concentrations of Solutions

103

3-6 CONCENTRATIONS OF SOLUTIONS Many chemical reactions are more conveniently carried out with the reactants mixed in solution rather than as pure substances. A solution is a homogeneous mixture, at the molecular level, of two or more substances. Simple solutions usually consist of one substance, the solute, dissolved in another substance, the solvent. The solutions used in the laboratory are usually liquids, and the solvent is often water. These are called aqueous solutions. For example, solutions of hydrochloric acid can be prepared by dissolving hydrogen chloride (HCl, a gas at room temperature and atmospheric pressure) in water. Solutions of sodium hydroxide are prepared by dissolving solid NaOH in water. We often use solutions to supply the reactants for chemical reactions. Solutions allow the most intimate mixing of the reacting substances at the molecular level, much more than would be possible in solid form. (A practical example is drain cleaner, shown in the photo.) We sometimes adjust the concentrations of solutions to speed up or slow down the rate of a reaction. In this section we study methods for expressing the quantities of the various components present in a given amount of solution. Concentrations of solutions are expressed in terms of either the amount of solute present in a given mass or volume of solution, or the amount of solute dissolved in a given mass or volume of solvent.

Percent by Mass Concentrations of solutions may be expressed in terms of percent by mass of solute, which gives the mass of solute per 100 mass units of solution. The gram is the usual mass unit. mass of solute percent solute ⫽ ᎏᎏ ⫻ 100% mass of solution mass of solute percent ⫽ ᎏᎏᎏᎏ ⫻ 100% mass of solute ⫹ mass of solvent

In some solutions, such as a nearly equal mixture of ethyl alcohol and water, the distinction between solute and solvent is arbitrary.

The sodium hydroxide and aluminum in some drain cleaners do not react while they are stored in solid form. When water is added, the NaOH dissolves and begins to act on trapped grease. At the same time, NaOH and Al react to produce H2 gas; the resulting turbulence helps to dislodge the blockage. Do you see why the container should be kept tightly closed?

Thus, a solution that is 10.0% calcium gluconate, Ca(C6H11O7)2, by mass contains 10.0 grams of calcium gluconate in 100.0 grams of solution. This could be described as 10.0 grams of calcium gluconate in 90.0 grams of water. The density of a 10.0% solution of calcium gluconate is 1.07 g/mL, so 100 mL of a 10.0% solution of calcium gluconate has a mass of 107 grams. Observe that 100 grams of a solution usually does not occupy 100 mL. Unless otherwise specified, percent means percent by mass, and water is the solvent.

EXAMPLE 3-13 Percent of Solute Calculate the mass of nickel(II) sulfate, NiSO4, contained in 200. g of a 6.00% solution of NiSO4. Plan The percentage information tells us that the solution contains 6.00 grams of NiSO4 per 100. grams of solution. The desired information is the mass of NiSO4 in 200. grams of solution. A unit factor is constructed by placing 6.00 grams NiSO4 over 100. grams of solution.

A 10.0% solution of Ca(C6H11O7)2 is sometimes administered intravenously in emergency treatment for black widow spider bites.

CHAPTER 3: Chemical Equations and Reaction Stoichiometry

Multiplication of the mass of the solution, 200. grams, by this unit factor gives the mass of NiSO4 in the solution. Solution 6.00 g NiSO4 __ ? g NiSO4 ⫽ 200. g soln ⫻ ᎏᎏ ⫽ 12.0 g NiSO4 100. g soln

EXAMPLE 3-14 Mass of Solution A 6.00% NiSO4 solution contains 40.0 g of NiSO4. Calculate the mass of the solution. Plan Placing 100. g of solution over 6.00 g of NiSO4 gives the desired unit factor. Solution 100. g soln __ ? g soln ⫽ 40.0 g NiSO4 ⫻ ᎏᎏ ⫽ 667 g soln 6.00 g NiSO4

EXAMPLE 3-15 Mass of Solute Calculate the mass of NiSO4 present in 200. mL of a 6.00% solution of NiSO4. The density of the solution is 1.06 g/mL at 25°C. Plan The volume of a solution multiplied by its density gives the mass of solution (see Section 1-11). The mass of solution is then multiplied by the mass fraction due to NiSO4 (6.00 g NiSO4/100. g soln) to give the mass of NiSO4 in 200. mL of solution. Solution

              

1.06 g soln 6.00 g NiSO4 __ ? g NiSO4 ⫽ 200. mL soln ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 12.7 g NiSO4 1.00 mL soln 100. g soln 212 g soln

You should now work Exercise 55.

EXAMPLE 3-16 Percent Solute and Density What volume of a solution that is 15.0% iron(III) nitrate contains 30.0 g of Fe(NO3)3? The density of the solution is 1.16 g/mL at 25°C. Plan Two unit factors relate mass of Fe(NO3)3 and mass of solution, 15.0 g Fe(NO3)3/100 g and 100 g/15.0 g Fe(NO3)3. The second factor converts grams of Fe(NO3)3 to grams of solution. Solution 100. g soln 1.00 mL soln __ ? mL soln ⫽ 30.0 g Fe(NO3)3 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 172 mL 15.0 g Fe(NO3)3 1.16 g soln

                  

104

200 g soln

3-6 Concentrations of Solutions

105

Note that the answer is not 200. mL but considerably less because 1.00 mL of solution has a mass of 1.16 grams; however, 172 mL of the solution has a mass of 200 g. You should now work Exercise 58.

See the Saunders Interactive

Molarity Molarity (M), or molar concentration, is a common unit for expressing the concentrations of solutions. Molarity is defined as the number of moles of solute per liter of solution: number of moles of solute molarity ⫽ ᎏᎏᎏᎏ number of liters of solution To prepare one liter of a one molar solution, one mole of solute is placed in a oneliter volumetric flask, enough solvent is added to dissolve the solute, and solvent is then added until the volume of the solution is exactly one liter. Students sometimes make the mistake of assuming that a one molar solution contains one mole of solute in a liter of solvent. This is not the case; one liter of solvent plus one mole of solute usually has a total volume of more than one liter. A 0.100 M solution contains 0.100 mol of solute per liter of solution, and a 0.0100 M solution contains 0.0100 mol of solute per liter of solution (Figure 3-2).

(a)

(b)

(c)

Figure 3-2 Preparation of 0.0100 M solution of KMnO4, potassium permanganate. A 250.-mL sample of 0.0100 M KMnO4 solution contains 0.395 g of KMnO4 (1 mol ⫽ 158 g). (a) 0.395 g of KMnO4 (0.00250 mol) is weighed out carefully and transferred into a 250.-mL volumetric flask. (b) The KMnO4 is dissolved in water. (c) Distilled H2O is added to the volumetric flask until the volume of solution is 250. mL. The flask is then stoppered, and its contents are mixed thoroughly to give a homogeneous solution.

General Chemistry CD-ROM, Screen 5.11, Preparing Solutions, Direct Addition.

The definition of molarity specifies the amount of solute per unit volume of solution, whereas percent specifies the amount of solute per unit mass of solution. Molarity therefore depends on temperature and pressure, whereas percent by mass does not.

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Water is the solvent in most of the solutions that we encounter. Unless otherwise indicated, we assume that water is the solvent. When the solvent is other than water, we state this explicitly.

EXAMPLE 3-17 Molarity Calculate the molarity (M) of a solution that contains 3.65 grams of HCl in 2.00 liters of solution. Plan We are given the number of grams of HCl in 2.00 liters of solution. We apply the definition of molarity, remembering to convert grams of HCl to moles of HCl. Solution __ ? mol HCl 3.65 g HCl 1 mol HCl ᎏᎏ ⫽ ᎏᎏ ⫻ ᎏᎏ ⫽ 0.0500 mol HCl/L soln L soln 2.00 L soln 36.5 g HCl

We place 3.65 g HCl over 2.00 L of solution, and then convert g HCl to mol HCl.

The concentration of the HCl solution is 0.0500 molar, and the solution is called 0.0500 M hydrochloric acid. One liter of the solution contains 0.0500 mol of HCl. You should now work Exercise 60.

EXAMPLE 3-18 Mass of Solute Calculate the mass of Ba(OH)2 required to prepare 2.50 L of a 0.0600 M solution of barium hydroxide. Plan The volume of the solution, 2.50 L, is multiplied by the concentration, 0.0600 mol Ba(OH)2/L, to give the number of moles of Ba(OH)2. The number of moles of Ba(OH)2 is then multiplied by the mass of Ba(OH)2 in one mole, 171.3 g Ba(OH)2/mol Ba(OH)2, to give the mass of Ba(OH)2 in the solution. Solution 171.3 g Ba(OH)2 0.0600 mol Ba(OH)2 __ ? g Ba(OH)2 ⫽ 2.50 L soln ⫻ ᎏᎏᎏ ⫻ ᎏᎏ 1 L soln 1 mol Ba(OH)2 ⫽ 25.7 g Ba(OH)2 You should now work Exercise 62.

The solutions of acids and bases that are sold commercially are too concentrated for most laboratory uses. We often dilute these solutions before we use them. We must know the molar concentration of a stock solution before it is diluted. This can be calculated from the specific gravity and the percentage data given on the label of the bottle.

EXAMPLE 3-19 Molarity A sample of commercial sulfuric acid is 96.4% H2SO4 by mass, and its specific gravity is 1.84. Calculate the molarity of this sulfuric acid solution.

3-7 Dilution of Solutions

Plan The density of a solution, grams per milliliter, is numerically equal to its specific gravity, so the density of the solution is 1.84 g/mL. The solution is 96.4% H2SO4 by mass; therefore 100. g of solution contains 96.4 g of pure H2SO4. From this information, we can find the molarity of the solution. First, we calculate the mass of one liter of solution. Solution __ ? g soln 1.84 g soln 1000 mL soln ᎏᎏ ⫽ ᎏᎏ ⫻ ᎏᎏ ⫽ 1.84 ⫻ 103 g soln/L soln L soln mL soln L soln The solution is 96.4% H2SO4 by mass, so the mass of H2SO4 in one liter is __ ? g H2SO4 1.84 ⫻ 103 g soln 96.4 g H2SO4 ᎏᎏ ⫽ ᎏᎏ ⫻ ᎏᎏ ⫽ 1.77 ⫻ 103 g H2SO4/L soln L soln L soln 100.0 g soln The molarity is the number of moles of H2SO4 per liter of solution. 1.77 ⫻ 103 g H2SO4 __ ? mol H2SO4 1 mol H2SO4 ᎏᎏ ⫽ ᎏᎏᎏ ⫻ ᎏᎏ ⫽ 18.0 mol H2SO4/L soln L soln L soln 98.1 g H2SO4 Thus, the solution is an 18.0 M H2SO4 solution. This problem can also be solved by using a series of three unit factors.

107

ANALYSIS Assay (H2SO4) W/W...Min. 95.0%--Max. 98.0%

MAXIMUM LIMITS OF IMPURITIES Appearance........Passes A.C.S. Test Color (APHA)....................10 Max. Residue after Ignition............4 ppm Chloride (Cl)....................0.2 ppm Nitrate (NO3)...................0.5 ppm Ammonium (NH4).................1 ppm Substances Reducing KMnO4 (limit about 2ppm as SO2).........Passes A.C.S. Test

Arsenic (As)..................0.004 ppm Heavy Metals (as Pb)..........0.8 ppm Iron (Fe).........................0.2 ppm Mercury (Hg).......................5 ppb Specific Gravity...................~1.84 Normality.............................~36 Suitable for Mercury Determinations A label that shows the analysis of sulfuric acid.

1 mol H2SO4 __ ? mol H2SO4 1.84 g soln 1000 mL soln 96.4 g H2SO4 ᎏᎏ ⫽ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ L soln mL soln L soln 100 g soln 98.1 g H2SO4 ⫽ 18.1 mol H2SO4/L soln ⫽ 18.1 M H2SO4 You should now work Exercise 68.

The small difference is due to rounding.

Problem-Solving Tip: Write Complete Units A common pitfall is to write units that are not complete enough to be helpful. For in1.84 g stance, writing the density in Example 3-19 as just ᎏ doesn’t help us figure mL 1.84 g soln 1000 mL soln out the required conversions. It is much safer to write ᎏᎏ, ᎏᎏ, mL soln L soln 96.4 g H2SO4 ᎏᎏ, and so on. In Example 3-19, we have written complete units to help 100 g soln guide us through the problem.

3-7 DILUTION OF SOLUTIONS Recall that the definition of molarity is the number of moles of solute divided by the volume of the solution in liters: number of moles of solute molarity ⫽ ᎏᎏᎏᎏ number of liters of solution

Multiplying both sides of the equation by the volume, we obtain volume (in L) ⫻ molarity ⫽ number of moles of solute

See the Saunders Interactive General Chemistry CD-ROM, Screen 5.12, Preparing Solutions, Dilution.

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Multiplication of the volume of a solution, in liters, by its molar concentration gives the amount of solute in the solution.

A can of frozen orange juice contains a certain mass (or moles) of vitamin C. After the frozen contents of the can are diluted by addition of water, the amount of vitamin C in the resulting total amount of solution will be unchanged. The concentration, or amount per a selected volume, will be less in the final solution, however.

We could use any volume unit as long as we use the same unit on both sides of the equation. This relationship also applies when the concentration is changed by evaporating some solvent.

When we dilute a solution by mixing it with more solvent, the amount of solute present does not change. But the volume and the concentration of the solution do change. Because the same number of moles of solute is divided by a larger number of liters of solution, the molarity decreases. Using a subscript 1 to represent the original concentrated solution and a subscript 2 to represent the dilute solution, we obtain volume1 ⫻ molarity1 ⫽ number of moles of solute ⫽ volume2 ⫻ molarity2 or

V1M1 ⫽ V2 M2

(for dilution only)

This expression can be used to calculate any one of four quantities when the other three are known (Figure 3-3). We frequently need a certain volume of dilute solution of a given molarity for use in the laboratory, and we know the concentration of the initial solution available. Then we can calculate the amount of initial solution that must be used to make the dilute solution.

Dilution of a concentrated solution, especially of a strong acid or base, frequently liberates a great deal of heat. This can vaporize drops of water as they hit the concentrated solution and can cause dangerous spattering. As a safety precaution, concentrated solutions of acids or bases are always poured slowly into water, allowing the heat to be absorbed by the larger quantity of water. Calculations are usually simpler to visualize, however, by assuming that the water is added to the concentrated solution.

C A U T I O N !

EXAMPLE 3-20 Dilution How many milliliters of 18.0 M H2SO4 are required to prepare 1.00 L of a 0.900 M solution of H2SO4? Plan The volume (1.00 L) and molarity (0.900 M) of the final solution, as well as the molarity (18.0 M) of the original solution, are given. Therefore, the relationship V1M1 ⫽ V2M2 can be used, with subscript 1 for the initial acid solution and subscript 2 for the dilute solution. We solve V1M1 ⫽ V2M2

for V1

Solution 1.00 L ⫻ 0.900 M V2M2 V1 ⫽ ᎏ ⫽ ᎏᎏᎏ ⫽ 0.0500 L ⫽ 50.0 mL 18.0 M M1

3-7 Dilution of Solutions

The dilute solution contains 1.00 L ⫻ 0.900 M ⫽ 0.900 mol of H2SO4, so 0.900 mol of H2SO4 must also be present in the original concentrated solution. Indeed, 0.0500 L ⫻ 18.0 M ⫽ 0.900 mol of H2SO4. You should now work Exercises 70 and 72.

(a)

(b)

K+ = CrO2– = 4

(c) Figure 3-3 Dilution of a solution. (a) A 100.-mL volumetric flask is filled to the calibration line with a 0.100 M potassium chromate, K2CrO4 solution. (b) The 0.100 M K2CrO4 solution is transferred into a 1.00-L volumetric flask. The smaller flask is rinsed with a small amount of distilled H2O. The rinse solution is added to the solution in the larger flask. To make sure that all the original K2CrO4 solution is transferred to the larger flask, the smaller flask is rinsed twice more and each rinse is added to the solution in the larger flask. (c) Distilled water is added to the 1.00-L flask until the liquid level coincides with its calibration line. The flask is stoppered and its contents are mixed thoroughly. The new solution is 0.0100 M K2CrO4. (100. mL of 0.100 M K2CrO4 solution has been diluted to 1000. mL.) The 100. mL of original solution and the 1000. mL of final solution both contain the amount of K2CrO4 dissolved in the original 100. mL of 0.100 M K2CrO4.

109

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3-8 USING SOLUTIONS IN CHEMICAL REACTIONS

See the Saunders Interactive General Chemistry CD-ROM, Screens 5.13 and 5.15, Stoichiometry of Reactions in Solution.

If we plan to carry out a reaction in a solution, we must calculate the amounts of solutions that we need. If we know the molarity of a solution, we can calculate the amount of solute contained in a specified volume of that solution. This procedure is illustrated in Example 3-21.

EXAMPLE 3-21 Amount of Solute Calculate (a) the number of moles of H2SO4 and (b) the number of grams of H2SO4 in 500. mL of 0.324 M H2SO4 solution. Plan Because we have two parallel calculations in this example, we state the plan for each step just before the calculation is done. 500. mL is more conveniently expressed as 0.500 L in this problem. By now, you should be able to convert mL to L (and the reverse) without writing out the conversion.

Solution (a) The volume of a solution in liters multiplied by its molarity gives the number of moles of solute, H2SO4 in this case. 0.324 mol H2SO4 __ ? mol H2SO4 ⫽ 0.500 L soln ⫻ ᎏᎏ ⫽ 0.162 mol H2SO4 L soln (b) We may use the results of part (a) to calculate the mass of H2SO4 in the solution.

A mole of H2SO4 is 98.1 g.

98.1 g H2SO4 __ ? g H2SO4 ⫽ 0.162 mol H2SO4 ⫻ ᎏᎏ ⫽ 15.9 g H2SO4 1 mol H2SO4 The mass of H2SO4 in the solution can be calculated without solving explicitly for the number of moles of H2SO4. 0.324 mol H2SO4 98.1 g H2SO4 __ ? g H2SO4 ⫽ 0.500 L soln ⫻ ᎏᎏ ⫻ ᎏᎏ ⫽ 15.9 g H2SO4 L soln 1 mol H2SO4

One of the most important uses of molarity relates the volume of a solution of known concentration of one reactant to the mass of the other reactant.

EXAMPLE 3-22 Solution Stoichiometry Calculate the volume in liters and in milliliters of a 0.324 M solution of sulfuric acid required to react completely with 2.792 grams of Na2CO3 according to the equation H2SO4 ⫹ Na2CO3 88n Na2SO4 ⫹ CO2 ⫹ H2O Plan The balanced equation tells us that one mole of H2SO4 reacts with one mole of Na2CO3, and we can write H2SO4 ⫹ Na2CO3 88n Na2SO4 ⫹ CO2 ⫹ H2O 1 mol

1 mol 106.0 g

1 mol

1 mol

1 mol

3-8 Using Solutions in Chemical Reactions

111

The indicator methyl orange changes from yellow, its color in basic solutions, to orange, its color in acidic solutions, when the reaction in Example 3-22 reaches completion.

We convert (1) grams of Na2CO3 to moles of Na2CO3, (2) moles of Na2CO3 to moles of H2SO4, and (3) moles of H2SO4 to liters of H2SO4 solution. g Na2CO3 88n mol Na2CO3 88n mol H2SO4 88n L H2SO4 soln Solution 1 L H2SO4 soln 1 mol Na2CO3 1 mol H2SO4 __ ? L H2SO4 ⫽ 2.792 g Na2CO3 ⫻ ᎏᎏ ⫻ ᎏᎏ ⫻ ᎏᎏ 106.0 g Na2CO3 1 mol Na2CO3 0.324 mol H2SO4 ⫽ 0.0813 L H2SO4 soln

or

81.3 mL H2SO4 soln

You should now work Exercise 76.

Often we must calculate the volume of solution of known molarity that is required to react with a specified volume of another solution. We always examine the balanced chemical equation for the reaction to determine the reaction ratio, that is, the relative numbers of moles of reactants.

EXAMPLE 3-23 Volume of Solution Required Find the volume in liters and in milliliters of a 0.505 M NaOH solution required to react with 40.0 mL of 0.505 M H2SO4 solution according to the reaction H2SO4 ⫹ 2NaOH 88n Na2SO4 ⫹ 2H2O Plan We shall work this example in several steps, stating the “plan,” or reasoning, just before each step in the calculation. Then we shall use a single setup to solve the problem.

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Solution The balanced equation tells us that the reaction ratio is 1 mol of H2SO4 to 2 mol of NaOH. H2SO4 ⫹ 2NaOH 88n Na2SO4 ⫹ 2H2O 1 mol

2 mol

1 mol

2 mol

From the volume and the molarity of the H2SO4 solution, we can calculate the number of moles of H2SO4. The volume of H2SO4 solution is expressed as 0.0400 L rather than 40.0 mL.

0.505 mol H2SO4 __ ? mol H2SO4 ⫽ 0.0400 L H2SO4 soln ⫻ ᎏᎏ ⫽ 0.0202 mol H2SO4 L soln The number of moles of H2SO4 is related to the number of moles of NaOH by the reaction ratio, 1 mol H2SO4/2 mol NaOH: 2 mol NaOH __ ? mol NaOH ⫽ 0.0202 mol H2SO4 ⫻ ᎏᎏ ⫽ 0.0404 mol NaOH 1 mol H2SO4 Now we can calculate the volume of 0.505 M NaOH solution that contains 0.0404 mol of NaOH:

Again we see that molarity is a unit factor. In this case, 1.00 L NaOH soln ᎏᎏᎏ 0.505 mol NaOH

1.00 L NaOH soln __ ? L NaOH soln ⫽ 0.0404 mol NaOH ⫻ ᎏᎏᎏ ⫽ 0.0800 L NaOH soln 0.505 mol NaOH which we usually call 80.0 mL of NaOH solution. We have worked through the problem stepwise; let us solve it in a single setup. L H2SO4 soln mol H2SO4 mol NaOH L NaOH 88n ᎏᎏ 88n ᎏᎏ ᎏᎏ 88n ᎏᎏ available available soln needed soln needed 2 mol NaOH 0.505 mol H2SO4 __ ? L NaOH soln ⫽ 0.0400 L H2SO4 soln ⫻ ᎏᎏ ⫻ ᎏᎏ L H2SO4 soln 1 mol H2SO4 1.00 L NaOH soln ⫻ ᎏᎏᎏ 0.505 mol NaOH ⫽ 0.0800 L NaOH soln or 80.0 mL NaOH soln You should now work Exercise 78.

Key Terms Actual yield The amount of a specified pure product actually obtained from a given reaction. Compare with Theoretical yield. Chemical equation Description of a chemical reaction by placing the formulas of reactants on the left and the formulas of products on the right of an arrow. A chemical equation must be balanced; that is, it must have the same number of each kind of atom on both sides. Concentration The amount of solute per unit volume or mass of solvent or of solution.

Dilution The process of reducing the concentration of a solute in solution, usually simply by adding more solvent. Limiting reactant A substance that stoichiometrically limits the amount of product(s) that can be formed. Molarity (M) The number of moles of solute per liter of solution. Percent by mass 100% multiplied by the mass of a solute divided by the mass of the solution in which it is contained. Percent yield 100% times actual yield divided by theoretical yield.

Exercises

Products Substances produced in a chemical reaction. Reactants Substances consumed in a chemical reaction. Reaction ratio The relative amounts of reactants and products involved in a reaction; may be the ratio of moles, or masses. Reaction stoichiometry Description of the quantitative relationships among substances as they participate in chemical reactions. Sequential reaction A chemical process in which several reaction steps are required to convert starting materials into products. Solute The dispersed (dissolved) phase of a solution.

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Solution A homogeneous mixture of two or more substances. Solvent The dispersing medium of a solution. Stoichiometry Description of the quantitative relationships among elements and compounds as they undergo chemical changes. Theoretical yield The maximum amount of a specified product that could be obtained from specified amounts of reactants, assuming complete consumption of the limiting reactant according to only one reaction and complete recovery of the product. Compare with Actual yield.

Exercises Chemical Equations 001. What is a chemical equation? What information does it contain? 002. When balancing chemical equations, you make certain that the same number of atoms of each element are on both sides of the equation. What scientific (natural) law requires that there be equal numbers of atoms of each element in both the products and the reactants? 003. Use words to state explicitly the relationships among numbers of molecules of reactants and products in the equation for the combustion of hexane, C6H14. 2C6H14 ⫹ 19O2 88n 12CO2 ⫹ 14H2O Balance each “equation” in Exercises 4–7 by inspection. 004. (a) Al ⫹ Cl2 88n Al2Cl6 (b) N2 ⫹ H2 88n NH3 (c) K ⫹ KNO3 88n K2O ⫹ N2 (d) H2O ⫹ KO2 88n KOH ⫹ O2 (e) H2SO4 ⫹ NH3 88n (NH4)2SO4 005. (a) P4 ⫹ O2 88n P4O6 (b) P4 ⫹ O2 88n P4O10 (c) K2CO3 ⫹ Al2Cl6 88n Al2(CO3)3 ⫹ KCl (d) KClO3 ⫹ C12H22O11 88n KCl ⫹ CO2 ⫹ H2O (e) KOH ⫹ H3PO4 88n KH2PO4 ⫹ H2O 006. (a) Fe2O3 ⫹ CO 88n Fe ⫹ CO2 (b) Mg3N2 ⫹ H2O 88n NH3 ⫹ Mg(OH)2 (c) Ca3(PO4)2 ⫹ H2SO4 88n Ca(H2PO4)2 ⫹ Ca(HSO4)2 (d) (NH4)2Cr2O7 88n N2 ⫹ H2O ⫹ Cr2O3 (e) Al ⫹ Cr2O3 88n Al2O3 ⫹ Cr 007. (a) UO2 ⫹ HF 88n UF4 ⫹ H2O (b) NaCl ⫹ H2O ⫹ SiO2 88n HCl ⫹ Na2SiO3 (c) Ca(HCO3)2 ⫹ Na2CO3 88n CaCO3 ⫹ NaHCO3 (d) NH3 ⫹ O2 88n NO ⫹ H2O (e) PCl3 ⫹ O2 88n POCl3

Calculations Based on Chemical Equations In Exercises 8–11, (a) write the balanced chemical equation that represents the reaction described by words, and then perform calculations to answer parts (b) and (c). 008. (a) Nitrogen, N2, combines with hydrogen, H2, to form ammonia, NH3. (b) How many hydrogen molecules are required to react with 600 nitrogen molecules? (c) How many ammonia molecules are formed in part (b)? 009. (a) Sulfur, S8, combines with oxygen at elevated temperatures to form sulfur dioxide. (b) If 250 oxygen molecules are used up in this reaction, how many sulfur molecules react? (c) How many sulfur dioxide molecules are formed in part (b)? 010. (a) Lime, CaO, dissolves in muriatic acid, HCl, to form calcium chloride, CaCl2, and water. (b) How many moles of HCl are required to dissolve 6.7 mol of CaO? (c) How many moles of water are formed in part (b)? 011. (a) Aluminum building materials have a hard, transparent, protective coating of aluminum oxide, Al2O3, formed by reaction with oxygen in the air. The sulfuric acid, H2SO4, in acid rain dissolves this protective coating and forms aluminum sulfate, Al2(SO4)3, and water. (b) How many moles of H2SO4 are required to react with 6.8 mol of Al2O3? (c) How many moles of Al2(SO4)3 are formed in part (b)? 012. Calculate the number of grams of baking soda, NaHCO3, that contain 14.0 moles of carbon. 013. Limestone, coral, and seashells are composed primarily of calcium carbonate. The test for the identification of a carbonate is to use a few drops of hydrochloric acid. The unbalanced equation is CaCO3 ⫹ HCl 88n CaCl2 ⫹ CO2 ⫹ H2O

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(a) Balance the equation. (b) How many atoms are in 0.250 moles of calcium carbonate? (c) What number of carbon dioxide molecules is released on the reaction of 0.250 moles of calcium carbonate? 014. How many moles of oxygen can be obtained by the decomposition of 8.0 mol of reactant in each of the following reactions? (a) 2KClO3 88n 2KCl ⫹ 3O2 (b) 2H2O2 88n 2H2O ⫹ O2 (c) 2HgO 88n 2Hg ⫹ O2 (d) 2NaNO3 88n 2NaNO2 ⫹ O2 (e) KClO4 88n KCl ⫹ 2O2 015. For the formation of 8.0 mol of water, which reaction uses the most nitric acid? (a) 3Cu ⫹ 8HNO3 88n 3Cu(NO3)2 ⫹ 2NO ⫹ 4H2O (b) Al2O3 ⫹ 6HNO3 88n 2Al(NO3)3 ⫹ 3H2O (c) 4Zn ⫹ 10HNO3 88n 4Zn(NO3)2 ⫹ NH4NO3 ⫹ 3H2O 016. Consider the reaction not balanced

NH3 ⫹ O2 8888888888n NO ⫹ H2O For every 25.00 mol of NH3, (a) how many moles of O2 are required, (b) how many moles of NO are produced, and (c) how many moles of H2O are produced? 017. What masses of cobalt(II) chloride and of hydrogen fluoride are needed to prepare 15.0 mol of cobalt(II) fluoride by the following reaction? CoCl2 ⫹ 2HF 88n CoF2 ⫹ 2HCl 018. We allow 24.0 g of methane, CH4, to react as completely as possible with excess oxygen, O2, to form CO2 and water. Write the balanced equation for this reaction. What mass of oxygen reacts? *19. Calculate the mass of calcium required to react with 3.770 g of carbon during the production of calcium carbide, CaC2. 020. Sodium iodide, NaI, is a source of iodine used to produce iodized salt. (a) Write the balanced chemical equation for the reaction of sodium and iodine. (b) How many grams of sodium iodide are produced by the reaction of 93.25 grams of iodine? 021. Consider the reaction

is condensed and found to weigh 18.75 g. Calculate the mass of Fe3O4 that reacted. 023. Iron(III) oxide, Fe2O3, is a result of the reaction of iron with the oxygen in air. (a) What is the balanced equation for this reaction? (b) What number of moles of iron react with 15.25 mol of oxygen from the air? (c) What mass of iron is required to react with 15.25 mol of oxygen? 024. Calculate the number of molecules of propane, C3H8, that will produce 4.80 grams of water when burned in excess oxygen, O2. 025. What mass of pentane, C5H12, produces 4.52 ⫻ 1022 CO2 molecules when burned in excess oxygen, O2?

Limiting Reactant 026. How many grams of NH3 can be prepared from 59.85 g of N2 and 12.11 g of H2? N2 ⫹ 3H2 88n 2NH3 *27. Silver nitrate solution reacts with calcium chloride solution according to the equation 2AgNO3 ⫹ CaCl2 88n Ca(NO3)2 ⫹ 2AgCl All of the substances involved in this reaction are soluble in water except silver chloride, AgCl, which forms a solid (precipitate) at the bottom of the flask. Suppose we mix together a solution containing 12.6 g of AgNO3 and one containing 8.40 g of CaCl2. What mass of AgCl is formed? *28. “Superphosphate,” a water-soluble fertilizer, is sometimes marketed as “triple phosphate.” It is a mixture of Ca(H2PO4)2 and CaSO4 on a 1:2 mole basis. It is formed by the reaction Ca3(PO4)2 ⫹ 2H2SO4 88n Ca(H2PO4)2 ⫹ 2CaSO4 We treat 300 g of Ca3(PO4)2 with 200 g of H2SO4. How many grams of superphosphate could be formed?

2NO ⫹ Br2 88n 2NOBr For every 7.50 mol of bromine that reacts, how many moles of (a) NO react and (b) NOBr are produced? 022. A sample of magnetic iron oxide, Fe3O4, reacts completely with hydrogen at red heat. The water vapor formed by the reaction heat

Fe3O4 ⫹ 4H2 8888n 3Fe ⫹ 4H2O

029. Gasoline is produced from crude oil, a nonrenewable resource. Ethanol is mixed with gasoline to produce a fuel called gasohol. Calculate the mass of water produced when

Exercises

66.89 g of ethanol, C2H5OH, is burned in 55.21 g of oxygen. 030. What mass of potassium can be produced by the reaction of 125.0 g of Na with 125.0 g of KCl? heat

Na ⫹ KCl 8888n NaCl ⫹ K 031. Silicon carbide, an abrasive, is made by the reaction of silicon dioxide with graphite. heat

SiO2 ⫹ C 8888n SiC ⫹ CO

(balanced?)

We mix 452 g of SiO2 and 306 g of C. If the reaction proceeds as far as possible, which reactant is left over? How much of this reactant remains? 032. Octane, C8H18, is a component of gasoline. A spark is used to ignite 1.563 grams of octane and 6.778 grams of oxygen in a sealed container. What mass of carbon dioxide is produced? 033. What mass of Ca(NO3)2 can be prepared by the reaction of 18.9 g of HNO3 with 7.4 g of Ca(OH)2? 2HNO3 ⫹ Ca(OH)2 88n Ca(NO3)2 ⫹ 2H2O 034. What is the maximum amount of Ca3(PO4)2 that can be prepared from 7.4 g of Ca(OH)2 and 9.8 g of H3PO4? 3Ca(OH)2 ⫹ 2H3PO4 88n Ca3(PO4)2 ⫹ 6 H2O 035. A reaction mixture contains 11.0 g of PCl3 and 7.00 g of PbF2. What mass of PbCl2 can be obtained from the following reaction? 3PbF2 ⫹ 2PCl3 88n 2PF3 ⫹ 3PbCl2 How much of which reactant is left unchanged?

Percent Yield from Chemical Reactions 036. The percent yield for the reaction PCl3 ⫹ Cl2 88n PCl5 is 83.2%. What mass of PCl5 is expected from the reaction of 73.7 g of PCl3 with excess chlorine? 037. The percent yield for the following reaction carried out in carbon tetrachloride solution is 59.0%.

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039. Tin(IV) chloride is produced in 85.0% yield by the reaction of tin with chlorine. How much tin is required to produce a kilogram of tin(IV) chloride? 040. Ethylene oxide, C2H4O, a fumigant sometimes used by exterminators, is synthesized in 88.1% yield by reaction of ethylene bromohydrin, C2H5OBr, with sodium hydroxide: C2H5OBr ⫹ NaOH 88n C2H4O ⫹ NaBr ⫹ H2O How many grams of ethylene bromohydrin are consumed in the production of 353 g of ethylene oxide, at 88.1% yield? 041. If 15.0 g sodium carbonate is obtained from the thermal decomposition of 75.0 g of sodium hydrogen carbonate, 2NaHCO3 88n Na2CO3 ⫹ H2O ⫹ CO2 what is the percent yield? 042. What is the percent yield if 112 mg SO2 is obtained from the combustion of 78.1 mg of carbon disulfide according to the reaction CS2 ⫹ 3O2 88n CO2 ⫹ 2SO2? 043. From a 60.0-g sample of an iron ore containing Fe3O4, 2.09 g of Fe is obtained by the reaction Fe3O4 ⫹ 2C 88n 3Fe ⫹ 2CO2 What is the percent of Fe3O4 in the ore? 044. The reaction of finely divided aluminum and iron(III) oxide, Fe2O3, is called the thermite reaction. It produces a tremendous amount of heat, making the welding of railroad track possible. The reaction of 750. grams of aluminum and 750. grams of iron(III) oxide produces 247.5 grams of iron. (a) Calculate the mass of iron that should be released by this reaction. (b) What is the percent yield of iron? Fe2O3 ⫹ 2Al 88n 2Fe ⫹ Al2O3 ⫹ heat 045. Lime, Ca(OH)2, can be used to neutralize an acid spill. A 5.06-g sample of Ca(OH)2 reacts with an excess of hydrochloric acid; 6.74 g of calcium chloride is collected. What is the percent yield of this experiment? Ca(OH)2 ⫹ 2HCl 88n CaCl2 ⫹ 2H2O

Br2 ⫹ Cl2 88n 2BrCl (a) What amount of BrCl is formed from the reaction of 0.0250 mol Br2 with 0.0250 mol Cl2? (b) What amount of Br2 is left unchanged? 038. Solid silver nitrate undergoes thermal decomposition to form silver metal, nitrogen dioxide, and oxygen. Write the chemical equation for this reaction. A 0.443-g sample of silver metal is obtained from the decomposition of a 0.784-g sample of AgNO3. What is the percent yield of the reaction?

Sequential Reactions 046. Consider the two-step process for the formation of tellurous acid described by the following equations: TeO2 ⫹ 2OH⫺ 88n TeO32⫺ ⫹ H2O TeO32⫺ ⫹ 2H⫹ 88n H2TeO3 What mass of H2TeO3 is formed from 95.2 g of TeO2, assuming 100% yield?

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047. Consider the formation of cyanogen, C2N2, and its subsequent decomposition in water given by the equations 2Cu2⫹ ⫹ 6CN⫺ 88n 2[Cu(CN)2]⫺ ⫹ C2N2 C2N2 ⫹ H2O 88n HCN ⫹ HOCN How much hydrocyanic acid, HCN, can be produced from 35.00 g of KCN, assuming 100% yield? 048. What mass of potassium chlorate is required to supply the proper amount of oxygen needed to burn 44.2 g of methane, CH4? 2KClO3 88n 2KCl ⫹ 3O2 CH4 ⫹ 2O2 88n CO2 ⫹ 2H2O 049. Hydrogen, obtained by the electrical decomposition of water, is combined with chlorine to produce 84.2 g of hydrogen chloride. Calculate the mass of water decomposed. 2H2O 88n 2H2 ⫹ O2 H2 ⫹ Cl2 88n 2HCl 050. Ammonium nitrate, known for its use in agriculture, can be produced from ammonia by the following sequence of reactions: NH3(g) ⫹ O2(g) 88n NO(g) ⫹ H2O(g) NO(g) ⫹ O2(g) 88n NO2(g) NO2(g) ⫹ H2O(ᐉ) 88n HNO3(aq) ⫹ NO(g) HNO3(aq) ⫹ NH3(g) 88n NH4NO3(aq) (a) Balance each equation. (b) How many moles of nitrogen atoms are required for every mole of ammonium nitrate (NH4NO3)? (c) How much ammonia is needed to prepare 100.0 grams of ammonium nitrate (NH4NO3)? 051. Calcium sulfate is the essential component of plaster and sheet rock. Waste calcium sulfate can be converted into quicklime, CaO, by reaction with carbon at high temperatures. The following two reactions represent a sequence of reactions that might take place: CaSO4(s) ⫹ 4C(s) 88n CaS(ᐉ) ⫹ 4CO(g) CaS(ᐉ) ⫹ 3CaSO4(s) 88n 4CaO(s) ⫹ 4SO2(g) What weight of sulfur dioxide (in grams) could be obtained from 1.000 kg of calcium sulfate? *52. The Grignard reaction is a two-step reaction used to prepare pure hydrocarbons. Consider the preparation of pure ethane, CH3CH3, from ethyl chloride, CH3CH2Cl. Step 1:

CH3CH2Cl ⫹ Mg 88n CH3CH2MgCl

Step 2: CH3CH2MgCl ⫹ H2O 88n CH3CH3 ⫹ Mg(OH)Cl

We allow 27.2 g of CH3CH2Cl (64.4 g/mol) to react with excess magnesium. From the first step reaction, CH3CH2MgCl (88.7 g/mol) is obtained in 79.5% yield. In the second step reaction, a 85.7% yield of CH3CH3 (30.0 g/mol) is obtained. What mass of CH3CH3 is obtained? *53. When sulfuric acid dissolves in water, the following reactions take place: H2SO4 88n H⫹ ⫹ HSO4⫺ HSO4⫺ 88n H⫹ ⫹ SO42⫺ The first reaction is 100.0% complete, and the second reaction is 10.0% complete. Calculate the concentrations of the various ions in a 0.150 M aqueous solution of H2SO4. *54. The chief ore of zinc is the sulfide, ZnS. The ore is concentrated by flotation and then heated in air, which converts the ZnS to ZnO. 2ZnS ⫹ 3O2 88n 2ZnO ⫹ 2SO2 The ZnO is then treated with dilute H2SO4 ZnO ⫹ H2SO4 88n ZnSO4 ⫹ H2O to produce an aqueous solution containing the zinc as ZnSO4. An electric current is passed through the solution to produce the metal. 2ZnSO4 ⫹ 2H2O 88n 2Zn ⫹ 2H2SO4 ⫹ O2 What mass of Zn is obtained from an ore containing 225 kg of ZnS? Assume the flotation process to be 90.6% efficient, the electrolysis step to be 92.2% efficient, and the other steps to be 100% efficient.

Concentrations of Solutions — Percent by Mass 055. (a) How many moles of solute are contained in 500 g of a 2.00% aqueous solution of K2Cr2O7? (b) How many grams of solute are contained in the solution of part (a)? (c) How many grams of water (the solvent) are contained in the solution of part (a)? 056. The density of an 18.0% solution of ammonium sulfate, (NH4)2SO4, is 1.10 g/mL. What mass of (NH4)2SO4 is required to prepare 275 mL of this solution? 057. The density of an 18.0% solution of ammonium chloride, NH4Cl, solution is 1.05 g/mL. What mass of NH4Cl does 275 mL of this solution contain? 058. What volume of the solution of (NH4)2SO4 described in Exercise 56 contains 90.0 g of (NH4)2SO4? *59. A reaction requires 33.6 g of NH4Cl. What volume of the solution described in Exercise 57 do you need if you wish to use a 25.0% excess of NH4Cl?

Exercises

Concentrations of Solutions — Molarity 060. What is the molarity of a solution that contains 555 g of phosphoric acid, H3PO4, in 3.00 L of solution? 061. What is the molarity of a solution that contains 4.50 g of sodium chloride in 40.0 mL of solution? 062. How many grams of the cleansing agent Na3PO4 (a) are needed to prepare 250 mL of 0.50 M solution, and (b) are in 250 mL of 0.50 M solution? 063. How many kilograms of ethylene glycol, C2H6O2, are needed to prepare a 9.50 M solution to protect a 14.0-L car radiator against freezing? What is the mass of C2H6O2 in 14.0 L of 9.50 M solution? 064. A solution made by dissolving 16.0 g of CaCl2 in 64.0 g of water has a density of 1.180 g/mL at 20°C. (a) What is the percent by mass of CaCl2 in the solution? (b) What is the molarity of CaCl2 in the solution? 065. A solution contains 0.100 mol/L of each of the following acids: HCl, H2SO4, H3PO4. (a) Is the molarity the same for each acid? (b) Is the number of molecules per liter the same for each acid? (c) Is the mass per liter the same for each acid? 066. What is the molarity of a barium chloride solution prepared by dissolving 1.50 g of BaCl2 ⭈ 2H2O in enough water to make 600 mL of solution? 067. How many grams of potassium benzoate trihydrate, KC7H5O2 ⭈ 3H2O, are needed to prepare 1 L of a 0.250 M solution of potassium benzoate? 068. Stock hydrofluoric acid solution is 49.0% HF and has a specific gravity of 1.17. What is the molarity of the solution? 069. Stock phosphoric acid solution is 85.0% H3PO4 and has a specific gravity of 1.70. What is the molarity of the solution?

Dilution of Solutions 070. Commercial concentrated hydrochloric acid is 12.0 M HCl. What volume of concentrated hydrochloric acid is required to prepare 4.50 L of 2.25 M HCl solution? 071. Commercially available concentrated sulfuric acid is 18.0 M H2SO4. Calculate the volume of concentrated sulfuric acid required to prepare 4.50 L of 2.25 M H2SO4 solution. 072. Calculate the volume of 0.0600 M Ba(OH)2 solution that contains the same number of moles of Ba(OH)2 as 225 mL of 0.0900 M Ba(OH)2 solution. 073. Calculate the volume of 4.00 M NaOH solution required to prepare 200 mL of a 0.800 M solution of NaOH. 074. Calculate the volume of pure water that dilutes 100. mL of 12 M NaOH to 4.75 M. 075. In the laboratory preparation room is a reagent bottle that contains 5.0 L of 12 M NaOH. Write a set of instructions

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for the production of 250 mL of 3.0 M NaOH from the 12 M solution.

Using Solutions in Chemical Reactions 076. Calculate the volume of a 0.225 M solution of potassium hydroxide, KOH, required to react with 0.215 g of acetic acid, CH3COOH, according to the following reaction. KOH ⫹ CH3COOH 88n KCH3COO ⫹ H2O 077. Calculate the number of grams of carbon dioxide, CO2, that can react with 135 mL of a 0.357 M solution of potassium hydroxide, KOH, according to the following reaction. 2KOH ⫹ CO2 88n K2CO3 ⫹ H2O 078. What volume of 0.246 M HNO3 solution is required to react completely with 38.6 mL of 0.0515 M Ba(OH)2? Ba(OH)2 ⫹ 2HNO3 88n Ba(NO3)2 ⫹ 2H2O 079. What volume of 0.55 M HBr is required to react completely with 0.80 mol of Ca(OH)2? 2HBr ⫹ Ca(OH)2 88n CaBr2 ⫹ 2H2O 080. An excess of AgNO3 reacts with 185.5 mL of an AlCl3 solution to give 0.325 g of AgCl. What is the concentration, in moles per liter, of the AlCl3 solution? AlCl3 ⫹ 3AgNO3 88n 3AgCl ⫹ Al(NO3)3 081. An impure sample of solid Na2CO3 is allowed to react with 0.1755 M HCl. Na2CO3 ⫹ 2HCl 88n 2NaCl ⫹ CO2 ⫹ H2O A 0.2337-g sample of sodium carbonate requires 15.55 mL of HCl solution. What is the purity of the sodium carbonate? 082. Calculate the volume of 12 M HCl that just reacts with 15 g of aluminum. 083. What volume of 18 M H2SO4 do you need to measure in order to neutralize 2 L of 5.35 M NaOH? The products are sodium sulfate and water. 084. What volume of 0.0496 M HClO4 reacts with 35.9 mL of a 0.505 M KOH solution according to the following reaction? KOH ⫹ HClO4 88n KClO4 ⫹ H2O *85. What is the molarity of a solution of sodium hydroxide, NaOH, if 36.9 mL of this solution is required to react with 29.2 mL of 0.101 M hydrochloric acid solution according to the following reaction? HCl ⫹ NaOH 88n NaCl ⫹ H2O

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*86. What is the molarity of a solution of sodium hydroxide, NaOH, if 29.8 mL of this solution is required to react with 25.0 mL of 0.0513 M nitric acid solution according to the following reaction? HNO3 ⫹ NaOH 88n NaNO3 ⫹ H2O

Mixed Exercises 087. What mass of sulfuric acid can be obtained from 1.00 kg of sulfur by the following series of reactions? 98% yield

S8 ⫹ 8O2 888888888n 8SO2 96% yield

2SO2 ⫹ O2 888888888n 2SO3 100% yield

SO3 ⫹ H2SO4 888888888n H2S2O7 97% yield

H2S2O7 ⫹ H2O 888888888n 2H2SO4 *88. What is the total mass of products formed when 38.8 g of carbon disulfide is burned in air? What mass of carbon disulfide must be burned to produce a mixture of carbon dioxide and sulfur dioxide that has a mass of 54.2 g? heat

CS2 ⫹ 3O2 8888n CO2 ⫹ 2SO2 *89. Iron(II) chloride, FeCl2, reacts with ammonia, NH3, and water, H2O, to produce iron(II) hydroxide, Fe(OH)2, and ammonium chloride, NH4Cl. (a) Write the balanced equation for this reaction. (b) We mix 78.5 g FeCl2, 25.0 g NH3, and 25.0 g H2O, which then react as completely as possible. Which is the limiting reactant? (c) How many grams of ammonium chloride, NH4Cl, are formed? (d) How many grams of each of the two leftover reactants remain at the completion of the reaction? *90. An iron ore that contains Fe3O4 reacts according to the reaction

093. Calculate the volume of 2.50 M phosphoric acid solution necessary to react with 45.0 mL of 0.150 M Mg(OH)2. 2H3PO4 ⫹ 3Mg(OH)2 88n Mg3(PO4)2 ⫹ 6H2O

CONCEPTUAL EXERCISES 094. Using your own words, give a definition of a chemical reaction. 095. Magnesium burns with a bright, white flame in air. (a) Write and balance the equation for the combination of magnesium and oxygen. (b) Explain the meaning of the numbers used to balance the equation. (c) How is a balanced equation an expression of the Law of Conservation of Matter? 096. A properly written and balanced chemical reaction is critical for the purposes of accurate communication. The oxidation of beryllium is proposed to follow the equation Be ⫹ O 88n BeO (a) Identify the basic error in the writing of this equation. (b) Provide the balanced equation. 097. How would you prepare 1 L of 1.25 ⫻ 10⫺6 M NaCl (molecular weight ⫽ 58.44 g/mol) solution by using a balance that can measure mass only to 0.01 g? 098. The drawings shown below represent beakers of aqueous solutions. Each sphere represents a dissolved solute particle. (a) Which solution is most concentrated? (b) Which solution is least concentrated? (c) Which two solutions have the same concentration? (d) When solutions E and F are combined, the resulting solution has the same concentration as solution ______.

Fe3O4 ⫹ 2C 88n 3Fe ⫹ 2CO2 We obtain 2.09 g of Fe from the reaction of 55.0 g of the ore. What is the percent Fe3O4 in the ore? *91. If 86.3% of the iron can be recovered from an ore that is 43.2% magnetic iron oxide, Fe3O4, what mass of iron could be recovered from 2.00 kg of this ore? The reduction of magnetic iron oxide is a complex process that can be represented in simplified form as

500. mL Solution A

500. mL Solution B

500. mL Solution C

500. mL Solution D

250. mL Solution E

250. mL Solution F

Fe3O4 ⫹ 4CO 88n 3Fe ⫹ 4CO2 092. Gaseous chlorine will displace bromide ion from an aqueous solution of potassium bromide to form aqueous potassium chloride and aqueous bromine. Write the chemical equation for this reaction. What mass of bromine is produced if 0.381 g of chlorine undergoes reaction?

Exercises

099. You prepared a NaCl solution by adding 58.44 g of NaCl to a 1-L volumetric flask and then adding water to dissolve it. When finished, the final volume in your flask looked like the illustration. The solution you prepared is (a) greater than 1 M because you added more solvent than necessary. (b) less than 1 M because you added less solvent than necessary. (c) greater than 1 M because you added less solvent than necessary. (d) less than 1 M because you added more solvent than necessary. (e) is 1 M because the amount of solute, not solvent, determines the concentration.

100. Zinc is more active chemically than is silver; it may be used to remove ionic silver from solution. Zn(s) ⫹ 2AgNO3(aq) 88n Zn(NO3)2(aq) ⫹ 2Ag(s) The concentration of a silver nitrate solution is determined to be 1.330 mol/L. Pieces of zinc totaling 100.0 g are added to 1.000 L of the solution; 90.0 g of silver is collected. (a) Calculate the percent yield of silver. (b) Suggest a reason why the yield is less than 100.0%. 101. Ammonia is formed in a direct reaction of nitrogen and hydrogen. N2(g) ⫹ 3H2(g) 88n 2NH3(g) A tiny portion of the starting mixture is represented by the diagram, where the blue spheres represent N and the white spheres represent H. Which of the following represents the product mixture?

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For the reaction of the given sample, which of the following is true? (a) N2 is the limiting reactant. (b) H2 is the limiting reactant. (c) NH3 is the limiting reactant. (d) No reactant is limiting; they are present in the correct stoichiometric ratio.

BUILDING YOUR KNOWLEDGE 102. Acetic acid, CH3COOH, reacts with ethanol, CH3CH2OH, to form ethyl acetate, CH3COOCH2CH3, (density ⫽ 0.902 g/mL) by the following reaction. CH3COOH ⫹ CH3CH2OH 88n CH3COOCH2CH3 ⫹ H2O We combine 20.2 mL of acetic acid with 20.1 mL of ethanol. (a) Which compound is the limiting reactant? (b) If 27.5 mL of pure ethyl acetate is produced, what is the percent yield? [Hint: See Tables 1-1 and 1-8.] 103. Concentrated hydrochloric acid solution is 37.0% HCl and has a density of 1.19 g/mL. A dilute solution of HCl

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CHAPTER 3: Chemical Equations and Reaction Stoichiometry

is prepared by diluting 4.50 mL of this concentrated HCl solution to 100.00 mL with water. Then 10.0 mL of this dilute HCl solution reacts with an AgNO3 solution according to the following reaction. HCl(aq) ⫹ AgNO3(aq) 88n HNO3(aq) ⫹ AgCl(s) How many milliliters of 0.108 M AgNO3 solution is required to precipitate all of the chloride as AgCl(s)? 104. In a particular experiment, 272 g of phosphorus, P4, reacted with excess oxygen to form tetraphosphorus decoxide, P4O10, in 89.5% yield. In the second step reaction, a 97.8% yield of H3PO4 was obtained. (a) Write the balanced equations for these two reaction steps. (b) What mass of H3PO4 was obtained? 105. Magnesium displaces copper from a dilute solution of copper(II) sulfate; the pure copper will settle out of the solution. Mg(s) ⫹ CuSO4(aq) 88n MgSO4(aq) ⫹ Cu(s) A copper(II) sulfate solution is mixed by dissolving 25.000 g of copper(II) sulfate, and then it is treated with an excess

of magnesium metal. The mass of copper collected is 8.786 g after drying. Calculate the percent yield of copper. 106. Suppose you are designing an experiment for the preparation of hydrogen. For the production of equal amounts of hydrogen, which metal, Zn or Al, is less expensive if Zn costs about half as much as Al on a mass basis? Zn ⫹ 2HCl 88n ZnCl2 ⫹ H2 2Al ⫹ 6HCl 88n 2AlCl3 ⫹ 3H2 107. Gaseous chlorine and gaseous fluorine undergo a combination reaction to form the interhalogen compound ClF. (a) Write the chemical equation for this reaction. (b) Calculate the mass of fluorine needed to react with 3.47 g of Cl2. (c) How many grams of ClF are formed? 108. It has been estimated that 93% of all atoms in the entire universe are hydrogen and that the vast majority of those remaining are helium. Based on only these two elements and the above-mentioned values, estimate the mass percentage composition of the universe.